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Express as a single logarithm with a coefficient of 1:

$$ 2(\ln(x)-\ln(x+1))-3(\ln(x^2)-\ln(x^2-1)) $$ I've been trying for nearly an hour and can't seem to find the answer, can anyone help plz :S

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up vote 3 down vote accepted

Looks like this what you are looking for: [first note that $\ln a+\ln b=\ln(ab)$, $\ln a-\ln b=\ln(a/b)$ and $x\ln a=\ln (a^x)$; assuming all the $\ln$'s exist]

Your expression $$=2\ln x+3\ln(x^2-1)-[2\ln(x+1)+3\ln(x^2)]$$ $$=\ln[x^2.(x^2-1)^3]-\ln[(x+1)^2.(x^2)^3]$$ $$=\ln\frac{x^2.(x^2-1)^3}{(x+1)^2.x^6}$$ $$=\ln (1+\frac{1}{x})(1-\frac{1}{x})^3$$

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Use the following rules for logarithms:

$$\ln(a)-\ln(b) = \ln\left(\frac{a}{b}\right)\\ c\cdot \ln(d) = \ln(d^c)$$

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Did you mean $c\ln(d)=\ln(d^c)$? – E.O. Sep 19 '12 at 4:22
@E.O. Whoops. Yes. I've edited to fix the error. – Richard Sullivan Sep 19 '12 at 12:03

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